## [sub-project] Perfect cuboid

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fwjmath
XBOX360-Installer Beiträge: 82
Registriert: 19.10.2010 15:26

### Re: [sub-project] Perfect cuboid

Zak hat geschrieben: You need give me an example to confirming your statement, otherwise mine is true and followed from Diophantine equations.

And you are making a big non-formal fallacy here. Even if I don't have a concrete example (I have, d=e=1, f=4), it does not mean that you are correct. It is like if you have a bogus proof of the Goldbach conjecture, I say that you are wroing, you ask me for a counter example of an even number impossible to be written as a sum of two primes, of course I cannot provide an example, but that does not make your "proof" correct.

You may want to learn above what a mathematical proof is.

Zak
Prozessor-Polier Beiträge: 109
Registriert: 04.11.2017 12:27

### Re: [sub-project] Perfect cuboid

fwjmath hat geschrieben:I have, d=e=1, f=4.
Let's insert yours values into expressions of this equation:
g ^ 2 = d ^ 2/2 + e ^ 2/2 + f ^ 2/2
So.
d ^ 2/2=1^2/2=0.5 - is not odd, cause not corresponds to 2k+1, were k belongs Z https://en.wikipedia.org/wiki/Parity_(mathematics)
It is followed that your statement "Not really. It is possible that exactly two of them are odd, and by adding them up we have an integer again." is ancorrect.
I think that we can't get odd value from expression type: d ^ 2/2, and if it is correct, so my statement is true.

fwjmath
XBOX360-Installer Beiträge: 82
Registriert: 19.10.2010 15:26

### Re: [sub-project] Perfect cuboid

Zak hat geschrieben:
fwjmath hat geschrieben:I have, d=e=1, f=4.
Let's insert yours values into expressions of this equation:
g ^ 2 = d ^ 2/2 + e ^ 2/2 + f ^ 2/2
So.
d ^ 2/2=1^2/2=0.5 - is not odd, cause not corresponds to 2k+1, were k belongs Z https://en.wikipedia.org/wiki/Parity_(mathematics)
It is followed that your statement "Not really. It is possible that exactly two of them are odd, and by adding them up we have an integer again." is ancorrect.
I think that we can't get odd value from expression type: d ^ 2/2, and if it is correct, so my statement is true.
That is exactly what I meant: it is possible that d and e are odd, and we will have d^2 / 2 + e^2 / 2 an integer. I don't mind if we can get integer value for d^2 / 2, as long as I have a solution for the original equation.

Can you please show that why d^2 / 2 must be an integer?

Zak
Prozessor-Polier Beiträge: 109
Registriert: 04.11.2017 12:27

### Re: [sub-project] Perfect cuboid

Zak hat geschrieben:O.K.
So. We have an equation:
g ^ 2 = d ^ 2/2 + e ^ 2/2 + f ^ 2/2
we may see that every summand in amount of right side must give an integger solution by dividing on 2, if we want to get integger g^2.
Thus we have a sum of three even values equal to square. So sqrt of sum of 3 even values will give an irrational solution.
I mean that sum of 3 even values is: d ^ 2/2 + e ^ 2/2 + f ^ 2/2, where I concider each summand like an expressions (d ^ 2/2), but not like variables(d,e,f). So each summand (expression) may be even, and there are a sum of 3 even values that gives irrational solution under sqrt.
Each summand (expression) may not to be odd, how you mean, because odd is always integger (Wiki)
But d ^ 2/2 may be integger when it is even, for example: 4^2/2=8. It's integer, because it's even.

fwjmath
XBOX360-Installer Beiträge: 82
Registriert: 19.10.2010 15:26

### Re: [sub-project] Perfect cuboid

Zak hat geschrieben:
Zak hat geschrieben:O.K.
So. We have an equation:
g ^ 2 = d ^ 2/2 + e ^ 2/2 + f ^ 2/2
we may see that every summand in amount of right side must give an integger solution by dividing on 2, if we want to get integger g^2.
Thus we have a sum of three even values equal to square. So sqrt of sum of 3 even values will give an irrational solution.
I mean that sum of 3 even values is: d ^ 2/2 + e ^ 2/2 + f ^ 2/2, where I concider each summand like an expressions (d ^ 2/2), but not like variables(d,e,f). So each summand (expression) may be even, and there are a sum of 3 even values that gives irrational solution under sqrt.
Each summand (expression) may not to be odd, how you mean, because odd is always integger (Wiki)
But d ^ 2/2 may be integger when it is even, for example: 4^2/2=8. It's integer, because it's even.
You still did not explain why d^2 / 2 must be an integer. If they can be non-integer (as I showed above), then your argument falls apart.

Zak
Prozessor-Polier Beiträge: 109
Registriert: 04.11.2017 12:27

### Re: [sub-project] Perfect cuboid

1. because d^2 / 2 never will odd, thus d^2 / 2 will be even and integger, it is followed from d - is integger value (face diagonal)
2. because g ^ 2 = d ^ 2/2 + e ^ 2/2 + f ^ 2/2 here, if we have a non-integger summ, we get non-integger value after sqrt action
But we always will have a sum of 3 even values under sqrt and irrational solutions.

fwjmath
XBOX360-Installer Beiträge: 82
Registriert: 19.10.2010 15:26

### Re: [sub-project] Perfect cuboid

Zak hat geschrieben:1. because d^2 / 2 never will odd, thus d^2 / 2 will be even and integger, it is followed from d - is integger value (face diagonal)
It does not follow. d^2 / 2 can be non-integer, which is neither odd nor even.
Zak hat geschrieben:2. because g ^ 2 = d ^ 2/2 + e ^ 2/2 + f ^ 2/2 here, if we have a non-integger summ, we get non-integger value after sqrt action
But we always will have a sum of 3 even values under sqrt and irrational solutions.
The sum must be an integer, but nothing says that each term must be an integer. It is possible that d^2 / 2 and e^2 / 2 are not integer, but their sum is an integer.

x3mEn
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Registriert: 20.03.2011 22:23

### Re: [sub-project] Perfect cuboid

2 * 1105^2 = 776^2 + 943^2 + 975^2

Zak
Prozessor-Polier Beiträge: 109
Registriert: 04.11.2017 12:27

### Re: [sub-project] Perfect cuboid

I came to this conclusion:
g ^ 2 = d ^ 2/2 + e ^ 2/2 + f ^ 2/2
if we have a sum of three even values equal to square then sqrt of sum of 3 even values will give an irrational solution thus d ^ 2/2, e ^ 2/2, f ^ 2/2 not may to accept event value each of them, it's clear.
g ^ 2, e ^ 2, f ^ 2 - integgers, it's clear to.
So is obtained that only one case have a place, when one of summand (f ^ 2/2) even and two others (d ^ 2/2, e ^ 2/2) - nither odd nor even. Thus we have d ^ 2/2 and e ^ 2/2 are with remainder of dividing 0.5 each, therefore d ^ 2 and e ^ 2 may have only odd values and f ^ 2/2 - only even. How do you think? Is it right?

fwjmath
XBOX360-Installer Beiträge: 82
Registriert: 19.10.2010 15:26

### Re: [sub-project] Perfect cuboid

Zak hat geschrieben:I came to this conclusion:
g ^ 2 = d ^ 2/2 + e ^ 2/2 + f ^ 2/2
if we have a sum of three even values equal to square then sqrt of sum of 3 even values will give an irrational solution thus d ^ 2/2, e ^ 2/2, f ^ 2/2 not may to accept event value each of them, it's clear.
g ^ 2, e ^ 2, f ^ 2 - integgers, it's clear to.
So is obtained that only one case have a place, when one of summand (f ^ 2/2) even and two others (d ^ 2/2, e ^ 2/2) - nither odd nor even. Thus we have d ^ 2/2 and e ^ 2/2 are with remainder of dividing 0.5 each, therefore d ^ 2 and e ^ 2 may have only odd values and f ^ 2/2 - only even. How do you think? Is it right?
Of course not. Try d=e=2, f=8, g=6.

Zak
Prozessor-Polier Beiträge: 109
Registriert: 04.11.2017 12:27

### Re: [sub-project] Perfect cuboid

If we have two equal face diagonals then will be get a cube, seemingly? And irrational space diagonal. How did you get f=8?
What kind of figure is that: d=e=2 and f=8?
Two face diagonals may be equal only if they bigger than third both?

x3mEn
XBOX360-Installer Beiträge: 99
Registriert: 20.03.2011 22:23

### Re: [sub-project] Perfect cuboid

Try d = 1552 e=1886 f=1950 g=2210:
2 * 2210^2 = 1552^2 + 1886^2 + 1950^2
The equation is true, d<e<f<g and d, e, f are even